Question: You have found the following ages (in years) of all 6 zebras at your local zoo: $ 22,\enspace 21,\enspace 13,\enspace 19,\enspace 16,\enspace 18$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{22 + 21 + 13 + 19 + 16 + 18}{{6}} = {18.2\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $3.8$ years $14.44$ years $^2$ $21$ years $2.8$ years $7.84$ years $^2$ $13$ years $-5.2$ years $27.04$ years $^2$ $19$ years $0.8$ years $0.64$ years $^2$ $16$ years $-2.2$ years $4.84$ years $^2$ $18$ years $-0.2$ years $0.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{14.44} + {7.84} + {27.04} + {0.64} + {4.84} + {0.04}} {{6}} $ $ {\sigma^2} = \dfrac{{54.84}}{{6}} = {9.14\text{ years}^2} $ The average zebra at the zoo is 18.2 years old. The population variance is 9.14 years $^2$.